TITLE: Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
QUESTION [11 upvotes]: In my textbook it asks for me to:
Prove that there is no constant $C$ such that $\text{arccot}(x) - \text{arctan}(\frac{1}{x}) = C $ for all $x \ne 0$. Explain why this does not violate the zero-derivative theorem.
But I believe I have found such a $C$, i.e. $C =0$!
I even asked WolframAlpha (http://www.wolframalpha.com/input/?i=arccot%28x%29+-+arctan%281%2Fx%29) which corroborates my answer.  
This question appears in Apostol's Calculus Volume I, Second Edition: Exersize 6.22-11b
Edit: Mathematica's definition of arccot is different from the one in my textbook. Apostol's arccot maps a real number into $(0, \pi)$ while Mathematica's maps a real number into $(-\pi/2, \pi/2)$ Here they are super-imposed: http://www.wolframalpha.com/input/?i=integral%28-1%2F%281%2Bx%5E2%29%29+%2B+pi%2F2%3B+arccot

REPLY [2 votes]: This is not a rigorous proof, it is intuitive one.
Let $y = \arctan\left(\dfrac1x\right)$ then 
$$
\begin{align}
\tan(y) &= \frac1x\\ 
\frac1{\tan(y)} &= x\\
\cot(y) &= x\\ 
y &= \text{arccot}(x)\\ 
\arctan\left(\frac1x\right) &= \text{arccot}(x)\\
\arctan\left(\frac1x\right) - \text{arccot}(x)&=0.
\end{align}
$$
It holds for $x \neq 0, \ x\in\mathbb{R}$.